_1":2,"_3":-5,"_4":-5},"loaderData",{"_5":6,"_7":8,"_9":-5,"_10":11,"_12":13,"_14":15,"_16":17,"_18":19,"_20":21},"actionData","errors","root",{"_2540":2541,"_2542":2543,"_2544":2545,"_2546":49,"_2547":2548,"_2549":2550,"_2551":49,"_2552":2553,"_2554":2555,"_2556":47,"_2557":2558,"_2559":-5,"_2560":-5,"_2561":-5,"_2562":-5,"_2563":-5,"_2564":-5,"_2565":2566,"_2567":2568,"_2569":49,"_2570":-5,"_119":49},"routes/__learning-environment",{"_640":2106,"_2107":2108,"_2109":-5,"_2110":2111,"_119":49},"routes/__learning-environment/course","routes/__learning-environment/course/$courseSlug",{"_598":599,"_600":601,"_602":49,"_603":153,"_119":49},"routes/__learning-environment/course/$courseSlug/books",{"_597":47,"_119":49},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug",{"_111":112,"_113":114,"_115":47,"_116":117,"_118":49,"_119":49},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug/chapters.$chapterSlug",{"_103":-5,"_107":108,"_109":110},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug/chapters.$chapterSlug/sections.$sectionSlug",{"_103":-5,"_104":49,"_105":106,"_22":23},"routes/__learning-environment/course/$courseSlug/books/$methodSlug.$bookSlug/chapters.$chapterSlug/sections.$sectionSlug/__section-content/faq",{"_22":23,"_24":25,"_26":27},"sectionTitle","9.4 Ongelijkheden oplossen","sectionBaseUrl","/course/wiskunde/books/moderne-wiskunde/leerboek-b-vwo-12/chapters/3-lineaire-vergelijkingen/sections/5-ongelijkheden-oplossen","questions",[28,29,30,31,32,33,34],{"_35":96,"_37":97,"_39":98,"_41":99,"_43":98,"_44":100,"_46":47,"_48":49,"_50":51,"_52":51,"_53":101,"_55":102},{"_35":88,"_37":89,"_39":90,"_41":91,"_43":92,"_44":93,"_46":47,"_48":49,"_50":51,"_52":51,"_53":94,"_55":95},{"_35":80,"_37":81,"_39":82,"_41":83,"_43":84,"_44":85,"_46":47,"_48":49,"_50":51,"_52":51,"_53":86,"_55":87},{"_35":73,"_37":74,"_39":75,"_41":76,"_43":75,"_44":77,"_46":47,"_48":49,"_50":51,"_52":51,"_53":78,"_55":79},{"_35":65,"_37":66,"_39":67,"_41":68,"_43":69,"_44":70,"_46":47,"_48":49,"_50":51,"_52":51,"_53":71,"_55":72},{"_35":57,"_37":58,"_39":59,"_41":60,"_43":61,"_44":62,"_46":47,"_48":49,"_50":51,"_52":51,"_53":63,"_55":64},{"_35":36,"_37":38,"_39":40,"_41":42,"_43":40,"_44":45,"_46":47,"_48":49,"_50":51,"_52":51,"_53":54,"_55":56},"uuid","da0746a3-a968-4444-81d9-1ba047399478","slug","wat-is-het-verschil-tussen-een-ongelijkheid-en-een-vergelijking","question","Wat is het verschil tussen een ongelijkheid en een vergelijking?","cleanedQuestion","verschil tussen ongelijkheid vergelijking","detailedQuestion","answer","Een ongelijkheid en een vergelijking zijn beide wiskundige uitdrukkingen, maar ze hebben een fundamenteel verschil in wat ze aangeven en wat voor soort oplossing je zoekt.\n\n### Wat is een vergelijking?\nEen **vergelijking** is een wiskundige uitdrukking waarbij je twee zijden aan elkaar gelijkstelt met een is-teken ($=$). Bij een vergelijking zoek je naar de specifieke waarde(n) van de variabele(n) waarvoor de linker- en rechterkant precies gelijk zijn. De oplossing van een vergelijking is meestal één of enkele specifieke getallen.\n* **Voorbeeld:** $2x + 1 = 7$\n Hier zoek je de exacte waarde van $x$ waarvoor de uitdrukking klopt.\n $2x = 6$\n $x = 3$\n De oplossing is één specifiek punt: $x=3$.\n\n### Wat is een ongelijkheid?\nEen **ongelijkheid** is een wiskundige uitdrukking waarbij je twee functies of expressies met elkaar vergelijkt met behulp van een ongelijkheidsteken. Je zoekt hier niet naar een exact gelijk punt, maar naar een bereik van waarden waarvoor de ene uitdrukking groter of kleiner is dan de andere.\nDe tekens die hiervoor worden gebruikt zijn:\n* $>$ (groter dan)\n* $<$ (kleiner dan)\n* $\\ge$ (groter dan of gelijk aan)\n* $\\le$ (kleiner dan of gelijk aan)\nDe oplossing van een ongelijkheid is een *bereik* van waarden, bijvoorbeeld $x < 2$ of $x \\ge -5$.\n\n### Hoe los je een ongelijkheid op en wat is de rol van het '=' teken?\nWanneer je een ongelijkheid oplost, is het vaak handig om eerst te bepalen waar de twee zijden van de ongelijkheid precies aan elkaar **gelijk** zijn. Dit punt, het snijpunt van de grafieken van de twee uitdrukkingen, is een cruciaal moment. Op dit punt verandert de relatie tussen de twee zijden: de ene kant gaat van 'groter dan' naar 'kleiner dan' of andersom.\n\nJe gebruikt het '=' teken dus om dat specifieke snijpunt te vinden, maar de oorspronkelijke vraag blijft een ongelijkheid. Je behandelt het even 'als een soort vergelijking' om dat ene punt te vinden, maar de uiteindelijke oplossing van de ongelijkheid is een *bereik* van waarden, en niet één specifiek getal.\n\n**Voorbeeld van het oplossen van een ongelijkheid:**\nStel je hebt de ongelijkheid $2x + 1 < -3x + 11$.\n\n1. **Vind het snijpunt (behandel als vergelijking):**\n $2x + 1 = -3x + 11$\n $5x = 10$\n $x = 2$\n Dit is het punt waar de twee uitdrukkingen gelijk zijn.\n\n2. **Bepaal de richting van de ongelijkheid:**\n * **Grafisch kijken:** Als je de grafieken van $y=2x+1$ en $y=-3x+11$ zou tekenen, zie je dat de grafiek van $y=2x+1$ onder de grafiek van $y=-3x+11$ ligt als $x < 2$.\n * **Algebraïsch (belangrijke regel):** Tijdens het algebraïsch oplossen van de ongelijkheid, behandel je het vaak net als een vergelijking, maar er is één cruciale regel:\n * Als je deelt of vermenigvuldigt met een **negatief getal**, dan moet het ongelijkheidsteken **omklappen**!\n\n Laten we het voorbeeld afmaken:\n $2x + 1 < -3x + 11$\n $5x < 10$ (hier heb je $3x$ opgeteld, teken blijft hetzelfde)\n $x < 2$ (hier heb je gedeeld door $5$, een positief getal, dus teken blijft hetzelfde)\n De oplossing is $x < 2$.\n\n **Voorbeeld van omklappen van het teken:**\n Stel je hebt de ongelijkheid $-3x < 9$.\n Om $x$ te vinden, deel je beide kanten door $-3$. Omdat je deelt door een **negatief getal**, klapt het ongelijkheidsteken om!\n $x > \\frac{9}{-3}$\n $x > -3$\n\nSamenvattend: een vergelijking zoekt naar specifieke punten waar twee uitdrukkingen gelijk zijn, terwijl een ongelijkheid zoekt naar een bereik van waarden waar de ene uitdrukking groter of kleiner is dan de andere. Het '=' teken wordt gebruikt als een hulpmiddel om het kritieke punt (snijpunt) te vinden bij het oplossen van een ongelijkheid.","isConcept",false,"isSynced",true,"likeCount",0,"dislikeCount","createdAt",["D",1790205623503],"updatedAt",["D",1790205623856],"0553f851-5ba2-4308-a8b4-a8278aaf4b53","hoe-bepaal-je-wanneer-lineaire-groei-groter-is-dan-ander","Hoe bepaal je in welk jaar lineaire groei groter is dan een andere?","bepaal welk jaar lineaire groei groter","Hoe reken ik bij lineaire groei uit in welk jaar het een groter is dan het ander?","Om bij lineaire groei uit te rekenen in welk jaar het ene groter is dan het andere, maak je gebruik van lineaire ongelijkheden. Dit betekent dat je twee lineaire functies met elkaar vergelijkt om te bepalen wanneer de waarde van de ene functie die van de andere overtreft.\n\nJe doorloopt hiervoor de volgende stappen:\n\n1. **Stel de ongelijkheid op:** Begin met het opstellen van een ongelijkheid die de situatie beschrijft. Dit houdt in dat je de twee lineaire groeiformules (bijvoorbeeld $y = ax + b$) tegenover elkaar zet met een ongelijkheidsteken ertussen. De veelgebruikte ongelijkheidstekens zijn:\n * $<$ (kleiner dan)\n * $>$ (groter dan)\n * $\\le$ (kleiner dan of gelijk aan)\n * $\\ge$ (groter dan of gelijk aan)\n\n *Voorbeeld:* Stel, je hebt twee situaties:\n * Situatie A: $y_A = 10x + 50$ (waarbij $x$ het aantal jaren is en $y_A$ de waarde van situatie A)\n * Situatie B: $y_B = 15x + 20$ (waarbij $x$ het aantal jaren is en $y_B$ de waarde van situatie B)\n Als je wilt weten wanneer situatie A groter is dan situatie B, stel je de ongelijkheid op: $10x + 50 > 15x + 20$.\n\n2. **Los de bijbehorende vergelijking op:** Om het 'omschakelpunt' te vinden, los je eerst de vergelijking op die hoort bij de ongelijkheid. Je vervangt het ongelijkheidsteken dus tijdelijk door een is-gelijk-teken (=). Dit snijpunt is het moment waarop de twee precies aan elkaar gelijk zijn.\n\n *Voorbeeld:* Voor de ongelijkheid $10x + 50 > 15x + 20$ los je eerst de vergelijking op:\n $10x + 50 = 15x + 20$\n $50 - 20 = 15x - 10x$\n $30 = 5x$\n $x = 6$\n Dit betekent dat na 6 jaar de waarden van situatie A en situatie B precies gelijk zijn.\n\n3. **Bepaal de oplossing van de ongelijkheid:** Nadat je het snijpunt hebt gevonden, kijk je wanneer de ene functie daadwerkelijk groter is dan de andere. Dit kun je doen door een waarde te testen die voor of na het snijpunt ligt, of door naar de grafieken van de functies te kijken.\n\n *Voorbeeld:* We weten dat bij $x=6$ de waarden gelijk zijn. We willen weten wanneer $10x + 50 > 15x + 20$.\n * Test een waarde kleiner dan 6, bijvoorbeeld $x=5$:\n * $y_A = 10(5) + 50 = 50 + 50 = 100$\n * $y_B = 15(5) + 20 = 75 + 20 = 95$\n Hier is $100 > 95$, dus $y_A > y_B$ als $x=5$.\n * Test een waarde groter dan 6, bijvoorbeeld $x=7$:\n * $y_A = 10(7) + 50 = 70 + 50 = 120$\n * $y_B = 15(7) + 20 = 105 + 20 = 125$\n Hier is $120 < 125$, dus $y_A < y_B$ als $x=7$.\n\n Uit de testwaarden blijkt dat situatie A groter is dan situatie B wanneer $x < 6$. Omdat $x$ vaak staat voor jaren en niet negatief kan zijn in dit soort contexten, is de oplossing $0 \\le x < 6$. Als de vraag specifiek is naar 'in welk jaar', dan zijn dit de jaren 0, 1, 2, 3, 4 en 5.\n\nDoor deze stappen te volgen, kun je bepalen in welke periode of welk jaar de ene lineaire groei groter is dan de andere.",["D",1779405373672],["D",1779405373904],"35e8e19e-f970-4ab2-aaf2-a37d8011ae0c","wanneer-draait-teken-om-bij-ongelijkheden","Hoe bepaal je de richting van het ongelijkheidsteken bij het oplossen van ongelijkheden?","Wanneer draait teken oplossen ongelijkheden","Hoe bepaal je de richting van het ongelijkheidsteken bij het oplossen van ongelijkheden, en wanneer draait het teken om?","Bij het oplossen van ongelijkheden is het cruciaal om te weten hoe je de juiste richting van het ongelijkheidsteken bepaalt. Dit betekent dat je moet vaststellen of de oplossing 'groter dan' ($>$), 'kleiner dan' ($<$), 'groter of gelijk aan' ($\\ge$) of 'kleiner of gelijk aan' ($\\le$) moet zijn.\n\nEr zijn twee belangrijke manieren om de richting van het ongelijkheidsteken te bepalen:\n\n1. **Grafisch kijken:**\n * Stel je voor dat je de grafieken van beide kanten van de ongelijkheid tekent.\n * Het snijpunt van deze grafieken is de $x$-waarde waarbij de twee kanten van de ongelijkheid precies gelijk zijn.\n * Kijk dan naar het oorspronkelijke ongelijkheidsteken.\n * Als er bijvoorbeeld staat $A < B$, dan zoek je naar de $x$-waarden waar de grafiek van $A$ **onder** de grafiek van $B$ ligt.\n * Als er staat $A > B$, dan zoek je waar de grafiek van $A$ **boven** de grafiek van $B$ ligt.\n * De 'of gelijk aan' tekens ($\\ge, \\le$) betekenen dat het snijpunt zelf ook meetelt in de oplossing.\n\n2. **Algebraïsch (let op de regels!):**\n * Tijdens het algebraïsch oplossen van de ongelijkheid, behandel je het vaak net als een vergelijking om het snijpunt te vinden.\n * **Heel belangrijk:** Het teken van een ongelijkheid draait om wanneer je beide kanten van de ongelijkheid vermenigvuldigt of deelt door een **negatief getal**. Dit is een cruciale regel om te onthouden, want als je dit niet doet, klopt je oplossing niet.\n\n**Waarom draait het teken om?**\nDenk aan een eenvoudige ongelijkheid: $2 < 5$. Dit is waar.\n* Als je beide kanten vermenigvuldigt met een positief getal, bijvoorbeeld 3:\n $2 \\times 3 < 5 \\times 3 \\Rightarrow 6 < 15$. De ongelijkheid blijft waar en het teken blijft hetzelfde.\n* Maar als je beide kanten vermenigvuldigt met een negatief getal, bijvoorbeeld -1:\n $2 \\times (-1)$ en $5 \\times (-1)$ worden $-2$ en $-5$.\n Nu is $-2$ *groter* dan $-5$. Dus, om de ongelijkheid waar te houden, moet het teken omdraaien: $-2 > -5$.\n\n**Voorbeeld:**\nStel je hebt de ongelijkheid $$-3x < 9$$.\nOm $x$ te isoleren, moet je beide kanten delen door $$-3$$. Omdat $$-3$$ een negatief getal is, moet je het ongelijkheidsteken omklappen.\n\nStap 1: Begin met de ongelijkheid:\n$$-3x < 9$$\n\nStap 2: Deel beide zijden door $$-3$$. Omdat je deelt door een negatief getal, draai je het ongelijkheidsteken om (van $<$ naar $>$):\n$$\\frac{-3x}{-3} > \\frac{9}{-3}$$\n\nStap 3: Vereenvoudig de uitdrukking:\n$$x > -3$$\n\nDe oplossing voor de ongelijkheid $$-3x < 9$$ is dus $$x > -3$$. Dit betekent dat elke waarde van $x$ die groter is dan $-3$ de oorspronkelijke ongelijkheid waar maakt.\n\n**Samenvattend:**\n* Als je deelt of vermenigvuldigt met een **positief getal**, blijft het ongelijkheidsteken hetzelfde.\n* Als je deelt of vermenigvuldigt met een **negatief getal**, klapt het ongelijkheidsteken om.\n\nHet snijpunt zelf blijft altijd de grens, of het nu positief of negatief is. Het is de richting van het ongelijkheidsteken die verandert als je met een negatief getal werkt.",["D",1777072735394],["D",1790205623088],"3c30e11a-a22a-462c-ba8b-d81e770a2a8a","ongelijkheden-getallenlijn-balansmethode","Hoe los ik ongelijkheden op met een getallenlijn en balansmethode?","los ongelijkheden getallenlijn balansmethode","Het oplossen van ongelijkheden met een getallenlijn en de balansmethode is een handige manier om de oplossing visueel te maken. De balansmethode wordt gebruikt om de 'scheidingspunten' te vinden, dit zijn de x-waarden waar de twee kanten van de ongelijkheid aan elkaar gelijk zijn. De getallenlijn helpt je vervolgens om te bepalen welke intervallen de oplossing vormen.\n\n### De balansmethode bij ongelijkheden\nDe balansmethode gebruik je om de kritieke punten te vinden. Dit zijn de x-waarden waar $f(x)$ en $g(x)$ aan elkaar gelijk zijn, oftewel waar de grafieken elkaar snijden.\n\nDe stappen zijn grotendeels hetzelfde als bij vergelijkingen:\n1. **Neem de ongelijkheid over.** Bijvoorbeeld: $f(x) > g(x)$.\n2. **Maak er eerst een vergelijking van:** $f(x) = g(x)$.\n3. **Los deze vergelijking op met de balansmethode** om de x-waarden te vinden waar $f(x)$ en $g(x)$ gelijk zijn. Dit zijn je 'kritieke' punten.\n\n**Belangrijk verschil bij ongelijkheden:**\nAls je tijdens het oplossen van een ongelijkheid beide kanten vermenigvuldigt of deelt door een **negatief getal**, moet je het ongelijkheidsteken omdraaien! Dit is cruciaal om de juiste oplossing te krijgen.\n\n### Met een getallenlijn\nZodra je de 'kritieke' x-waarden hebt gevonden (met de balansmethode), kun je deze gebruiken op een getallenlijn:\n1. **Teken een getallenlijn** en markeer hierop de gevonden 'kritieke' x-waarden. Deze punten verdelen de getallenlijn in verschillende intervallen.\n2. **Kies testpunten:** Kies in elk interval een willekeurige x-waarde (een 'testpunt') en vul deze in de oorspronkelijke ongelijkheid in.\n3. **Controleer de ongelijkheid:** Kijk of de ongelijkheid waar is voor dat testpunt. Als de ongelijkheid waar is, dan is het hele interval waar dat testpunt in ligt, onderdeel van de oplossing.\n4. **Markeer de oplossing:**\n * Als de ongelijkheid een strikt teken heeft ($<$ of $>$), gebruik je een **open bolletje** bij de kritieke punten op de getallenlijn, omdat deze punten zelf niet meedoen.\n * Als de ongelijkheid een 'gelijk aan' teken heeft ($\\le$ of $\\ge$), gebruik je een **dicht bolletje**, omdat de kritieke punten dan wel meedoen.\n * Kleur vervolgens de intervallen in die voldoen aan de ongelijkheid.",["D",1773447601292],["D",1773447601575],"d2501d80-24aa-432a-bfe4-27c59204d555","hoe-interpreteer-je-ongelijkheden-in-een-grafiek","Hoe interpreteer je ongelijkheden in een grafiek?","interpreteer ongelijkheden grafiek","Hoe interpreteer je ongelijkheden in een grafiek, en hoe bepaal je of een lijn 'boven' of 'onder' een andere lijn ligt om de oplossing te vinden?","Om ongelijkheden in een grafiek te interpreteren, kijk je naar de relatieve posities van de lijnen ten opzichte van elkaar. Dit helpt je te bepalen voor welke waarden van de variabele de ongelijkheid geldt.\n\nHier is hoe het werkt:\n\n1. **Begrijp de ongelijkheid:**\n * Als je een ongelijkheid hebt zoals $A > B$, betekent dit dat je zoekt naar de punten waar de grafiek van $A$ **boven** de grafiek van $B$ ligt.\n * Als je een ongelijkheid hebt zoals $A < B$, betekent dit dat je zoekt naar de punten waar de grafiek van $A$ **onder** de grafiek van $B$ ligt.\n\n2. **Vind het snijpunt:**\n Het snijpunt van de twee lijnen is cruciaal. Op dit punt zijn de waarden van beide functies precies gelijk ($A = B$). Dit punt verdeelt de grafiek in gebieden waar de ene functie groter is dan de andere.\n * **Voorbeeld:** Stel Sophia verkoopt ijsjes. Haar opbrengst ($R$) is $2.25q$ (waarbij $q$ het aantal verkochte ijsjes is) en haar kosten ($K$) zijn $1.15q + 352$. Ze maakt winst als haar opbrengst groter is dan haar kosten, dus $R > K$.\n * Om het snijpunt te vinden, los je de vergelijking $R = K$ op:\n $2.25q = 1.15q + 352$\n $1.1q = 352$\n $q = \\frac{352}{1.1}$\n $q = 320$\n * Bij $q = 320$ ijsjes zijn de opbrengst en de kosten gelijk (dit is het 'break-even point').\n\n3. **Interpreteer de grafiek:**\n Nadat je het snijpunt hebt gevonden, kijk je naar de grafiek om te zien waar de lijn van de ene functie boven of onder de lijn van de andere functie ligt.\n * In het voorbeeld van Sophia: we willen weten wanneer $R > K$. Dit betekent dat we zoeken naar het gebied waar de grafiek van de opbrengst ($R$) **boven** de grafiek van de kosten ($K$) ligt.\n * Als je naar de grafiek kijkt, zie je dat de lijn van $R$ boven de lijn van $K$ ligt als $q$ groter is dan het snijpunt ($q = 320$).\n * De conclusie is dan: Sophia moet meer dan 320 ijsjes verkopen om winst te maken ($q > 320$).\n\n**Lijnen tekenen om te interpreteren:**\nSoms moet je zelf de lijnen tekenen om de ongelijkheid te kunnen interpreteren. Om een rechte lijn te tekenen, heb je minimaal twee punten nodig. Je vindt deze punten door willekeurige waarden voor $x$ (of $y$) te kiezen en de bijbehorende $y$ (of $x$) te berekenen.\n\n* **Voorbeeld:** Om de lijn $2x - 5y = 8$ te tekenen:\n * **Kies $x = -1$:**\n $2(-1) - 5y = 8$\n $-2 - 5y = 8$\n $-5y = 10$\n $y = -2$\n Dit geeft het punt $(-1, -2)$.\n * **Kies $y = 0$:**\n $2x - 5(0) = 8$\n $2x = 8$\n $x = 4$\n Dit geeft het punt $(4, 0)$.\n * Met deze twee punten $(-1, -2)$ en $(4, 0)$ kun je de lijn tekenen en vervolgens ongelijkheden interpreteren die met deze lijn te maken hebben.\n\nDoor deze stappen te volgen, kun je ongelijkheden in een grafiek correct interpreteren en de oplossing bepalen.",["D",1775777004754],["D",1775777005029],"087e2ad2-1dd3-450f-ba8c-4f8b19861ae4","hoe-bereken-je-de-y-coordinaat-van-een-snijpunt","Hoe bereken je de y-coördinaat van een snijpunt?","bereken ycoördinaat snijpunt","Hoe bereken je de y-coördinaat van een snijpunt als je de x-coördinaat en de vergelijkingen van de lijnen weet?","Om de y-coördinaat van een snijpunt te berekenen, volg je deze stappen:\n\n1. **Vind de x-coördinaat:** Zorg ervoor dat je eerst de x-coördinaat van het snijpunt hebt berekend. Dit doe je meestal door de vergelijkingen van de twee lijnen aan elkaar gelijk te stellen en op te lossen voor $x$.\n2. **Vul de x-coördinaat in:** Kies één van de oorspronkelijke vergelijkingen van de lijnen. Het maakt niet uit welke je kiest, want op het snijpunt hebben beide vergelijkingen dezelfde $y$-waarde voor die specifieke $x$-waarde.\n3. **Bereken de y-waarde:** Vul de gevonden x-coördinaat in de gekozen vergelijking in en reken de bijbehorende y-waarde uit.\n\n**Voorbeeld:**\nStel, je hebt de x-coördinaat van een snijpunt gevonden, bijvoorbeeld $x = 1.5$.\nEén van de oorspronkelijke vergelijkingen van de lijnen is $y = 0.8x$.\nOm de y-coördinaat te vinden, vul je $x = 1.5$ in deze vergelijking in:\n$y = 0.8 \\times 1.5$\n$y = 1.2$\nDe y-coördinaat van het snijpunt is dus $1.2$.",["D",1775777005363],["D",1775777006312],"80e54b46-8112-4fee-9427-2cc5a49d22db","wat-betekent-een-ongelijkheid","Wat betekent een ongelijkheid?","betekent ongelijkheid","Een ongelijkheid is een wiskundige uitspraak die aangeeft wanneer de ene uitdrukking groter of kleiner is dan de andere, in plaats van precies gelijk. Waar je bij een gewone vergelijking (zoals $f(x) = g(x)$) zoekt wanneer twee uitdrukkingen precies gelijk aan elkaar zijn (denk aan de snijpunten van grafieken), kijk je bij een ongelijkheid naar wanneer de ene uitdrukking groter of kleiner is dan de andere.\n\nDe tekens die hiervoor worden gebruikt zijn:\n* $>$ (groter dan)\n* $<$ (kleiner dan)\n* $\\ge$ (groter dan of gelijk aan)\n* $\\le$ (kleiner dan of gelijk aan)\n\nAls je bijvoorbeeld $f(x) > g(x)$ ziet staan, dan zoek je naar alle $x$-waarden waarvoor de grafiek van $f$ boven de grafiek van $g$ ligt. 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